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\(\left|3x+4\right|=2\left|2x-9\right|\)
\(\left|3x+4\right|\ge0\)
\(\left|2x-9\right|\ge0\Rightarrow2\left|2x-9\right|\ge0\)
\(\Rightarrow3x+4=2\left(2x-9\right)\)
\(3x+4=4x-18\)
\(3x=4x-14\)
\(x=14\)
\(\left|10x+7\right|\le37\)
\(\Rightarrow\left|10x+7\right|\le\left\{37;36;35;......;0\right\}\)
\(10x+7\le\left\{\pm37;\pm36;\pm35;.....0\right\}\)
Tự tính tiếp.C tương tự
\(\left|x+3\right|-2x=\left|x-4\right|\)
\(\left|x+3\right|=\left|x-4\right|+2x\)
\(\left|x+3\right|\ge0\)
\(\left|x-4\right|\ge0\)
\(\Rightarrow x+3=x-4+2x\)
\(x+3=3x-4\)
\(x=3x-7\)
\(x=\dfrac{7}{2}\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
a) Ta có : \(\left|3x+4\right|=2\left|2x-9\right|\)
=> \(\orbr{\begin{cases}3x+4=2\left(-2x+9\right)\\3x+4=2\left(2x-9\right)\end{cases}}\Rightarrow\orbr{\begin{cases}3x+4=-4x+18\\3x+4=4x-18\end{cases}}\Rightarrow\orbr{\begin{cases}7x=14\\-x=-22\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=22\end{cases}}\)
=> \(x\in\left\{2;22\right\}\)
b) Ta có : \(\left|10x+7\right|< 37\)
=> -37 < 10x + 7 < 37
=> -44 < 10x < 30
=> -4,4 < x < 3
Vậy -4,4 < x < 3
c) |3 - 8x| \(\le\)19
=> \(-19\le3-8x\le19\)
=> \(\hept{\begin{cases}3-8x\ge-19\\3-8x\le19\end{cases}}\Rightarrow\hept{\begin{cases}22\ge8x\\-16\le8x\end{cases}}\Rightarrow\hept{\begin{cases}x\le\frac{11}{4}\\x\ge-2\end{cases}}\Rightarrow-2\le x\le\frac{11}{4}\)
d) Ta có |x + 3| - 2x = |x - 4| (1)
Nếu x < -3
=> |x + 3| = -(x + 3) = -x - 3
=> |x - 4| = -(x - 4) = -x + 4
Khi đó (1) <=> -x - 3 - 2x = - x + 4
=> -3x - 3 = - x + 4
=> -2x = 7
=> x = - 3,5 (tm)
Nếu \(-3\le x\le4\)
=> |x + 3| = x + 3
=> |x - 4| = -(x - 4) = -x + 4
Khi đó (1) <=> x + 3 - 2x = -x + 4
=> -x + 3 = -x + 4
=> 0x = 1 (loại)
Nếu x > 4
=> |x + 3| = x + 3
=> |x - 4| = x + 4
Khi đó (1) <=> x + 3 - 2x = x - 4
=> -x + 3 = x - 4
=> -2x = -7
=> x = 3,5 (loại)
Vậy x = -3,5